Question

Two boxes are connected by a rope strung over a pulley. Box A with a mass of 5.0 kg sits on the surface of a plane inclined with an angle of 33 degrees. The coefficient of static and kinetic friction between the box A and the surface it sits on is 0.32 and 0.20 respectively. The rope connecting Box A and Box B is looped over a frictionless pulley at the top of the incline and Box B hangs free in the air suspended below the pulley. If both boxes are motionless, what is the mass of Box B? If Box B is one kilogram heavier than that what is the acceleration of the boxes?

Answer 1

s = coeffcient of static friction = 0.32

k = coeffcient of kinetic friction = 0.20

M_a = mass of box A = 5 kg

M_b = mass of box B

for mass A :

force equation perpendicular to incline plane is given as

F_n = M_a g Cos

F_n = 5 x 9.8 Cos33 = 41.1 N

static friction force acting

f_s = s F_n = s M_a g Cos = 0.32 x 41.1 = 13.2 N

kinetic friction force

f_k = k F_n = k M_a g Cos = 0.20 x 41.1 = 8.22 N

force equation parallel to incline plane is given as

T = M_a g Sin + f_s

T = 5 x 9.8 Sin33 + 13.2

T = 39.88 N

for mass B :

force equation is given as

T = M_b g

39.88 = M_b (9.8)

M_b = 4.1 kg

ii)

new mass of Block B = M'_b = 4.1 + 1 = 5.1 kg

weight of box B = 5.1 x 9.8 = 49.98 N

force equation for the motion of block B

M_bg - T = M_ba

5.1 x 9.8 - T = 5.1a

49.98 - T = 5.1 a                   

T = 49.98 - 5.1 a eq-1

force equation for the motion of block A

T - M_a g Sin - f_k = M_a a

(49.98 - 5.1 a) - 5 x 9.8 Sin33 - 8.22 = 5a

a = 1.5 m/s²